Question

Cho \(a,b,c>0\), cmr: \(\dfrac{1}{a}+\dfrac{4}{b}+\dfrac{9}{c}\ge\dfrac{36}{a+b+c}\)(Dùng bđt Svác-xơ)

Answer 1

Lời giải:

Áp dụng BĐT Svac-xơ ta có:

\(\frac{1}{a}+\frac{4}{b}+\frac{9}{c}=\frac{1^2}{a}+\frac{2^2}{b}+\frac{3^2}{c}\geq \frac{(1+2+3)^2}{a+b+c}=\frac{36}{a+b+c}\)

Ta có đpcm

Dấu "=" xảy ra khi \(\frac{1}{a}=\frac{2}{b}=\frac{3}{c}\)